If f(x) = 2 ^4 x3 + 2 ^4 x + 2 ^4 x3 + 2 ^4 x, then 15 f'(0) is equal to:

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Given: f(x) is to be evaluated for 15 f'(0). From the solution working, use row operations on the determinant form of f(x). Find: The value of 15 f'(0). By simplifying the determinant using row operations, R2 R2 - R1, R3 R3 - R1 we find that f(x) is constant. Therefore, f'(x) = 0 so at x = 0, f'(0) = 0 Hence, 15 f'(0) = 0 Therefore, the correct option is A.

Given: The extracted solution rewrites the problem in determinant form and evaluates the derivative at $x=0$ .

Find: $\frac{1}{5} f'(0)$ .

The extracted solution states:

f(x)=\begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix}

At $x=0$ ,

\cos 0 = 1, \qquad \sin 0 = 0

so the matrix becomes

f(0)=\begin{vmatrix} 2 & 0 & 3 \\ 5 & 0 & 0 \\ 2 & 3 & 0 \end{vmatrix}

The extracted working then argues that after differentiation there is no linear variation at $x=0$ , hence

f'(0)=0

Therefore,

\frac{1}{5} f'(0)=\frac{1}{5}\times 0 = 0

So the correct option is A.

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